Đáp án:
$\begin{array}{l}
a)Dkxd:x \ge 0;x \ne 9\\
D = \left( {\dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x }}{{\sqrt x - 3}} - \dfrac{{3x + 3}}{{x + 9}}} \right):\left( {\dfrac{{2\sqrt x - 2}}{{\sqrt x - 3}} - 1} \right)\\
= \dfrac{{2\sqrt x .\left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
:\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\left( {\sqrt x + 1} \right)}}{{\sqrt x + 3}}.\dfrac{1}{{\sqrt x + 1}}\\
= \dfrac{{ - 3}}{{\sqrt x + 3}}\\
b)D = - 1\\
\Rightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} = - 1\\
\Rightarrow \sqrt x + 3 = 3\\
\Rightarrow \sqrt x = 0\\
\Rightarrow x = 0\left( {tmdk} \right)\\
\text{Vậy}\,x = 0\\
c)D = 16\\
\Rightarrow \dfrac{{ - 3}}{{\sqrt x + 3}} = 16\\
\Rightarrow \sqrt x + 3 = - \dfrac{3}{{16}}\\
\Rightarrow \sqrt x = \dfrac{{ - 51}}{{16}}\left( {ktm} \right)
\end{array}$
Vậy ko có x để D = 16
